22. (10分)如图,在$□ ABCD$中,过点$D$作$DE \perp BC$,垂足为$E$,连接$AE,F$为线段$AE$上一点,且$\angle DFE=\angle C$.
(1) 求证:$\frac{BC}{AE}=\frac{DF}{CD}$;
(2) 若$AB=10,AD=8\sqrt{3},DE=8$,求$DF$的长.
(1) 求证:$\frac{BC}{AE}=\frac{DF}{CD}$;
(2) 若$AB=10,AD=8\sqrt{3},DE=8$,求$DF$的长.
答案：22.(1)
∵四边形ABCD是平行四边形，
∴AD = BC,AB = CD,AB//CD,AD//BC.
∴∠B + ∠C = 180°,∠DAE = ∠AEB.
∵∠DFE = ∠C,∠AFD + ∠DFE = 180°,
∴∠B = ∠AFD.
∴△ADF∽△EAB.
∴$\frac{AD}{EA} = \frac{DF}{AB}.$
∴$\frac{BC}{AE} = \frac{DF}{CD}$
(2)
∵四边形ABCD是平行四边形，
∴AD//BC.
∵DE⊥BC,
∴DE⊥AD.
∴∠ADE = 90°.在Rt△ADE中,由勾股定理，得AE = √{AD² + DE²} = √{(8√{3})² + 8²} = 16.由(1)知,△ADF∽△EAB,
∴$\frac{AD}{EA} = \frac{DF}{AB}.$
∴$\frac{8√{3}}{16} = \frac{DF}{10},$解得DF = 5√{3}
∵四边形ABCD是平行四边形，
∴AD = BC,AB = CD,AB//CD,AD//BC.
∴∠B + ∠C = 180°,∠DAE = ∠AEB.
∵∠DFE = ∠C,∠AFD + ∠DFE = 180°,
∴∠B = ∠AFD.
∴△ADF∽△EAB.
∴$\frac{AD}{EA} = \frac{DF}{AB}.$
∴$\frac{BC}{AE} = \frac{DF}{CD}$
(2)
∵四边形ABCD是平行四边形，
∴AD//BC.
∵DE⊥BC,
∴DE⊥AD.
∴∠ADE = 90°.在Rt△ADE中,由勾股定理，得AE = √{AD² + DE²} = √{(8√{3})² + 8²} = 16.由(1)知,△ADF∽△EAB,
∴$\frac{AD}{EA} = \frac{DF}{AB}.$
∴$\frac{8√{3}}{16} = \frac{DF}{10},$解得DF = 5√{3}
23. (12分)如图,一次函数$y=k_1x+b(k_1 \neq 0)$的图象与反比例函数$y=\frac{k_2}{x}$的图象交于点$A(-1,6),B(a,2-\frac{12}{a})$,与$x$轴交于点$C$.
(1) 求反比例函数与一次函数的解析式;
(2) 若点$M$在$x$轴上$,S_{\triangle OAM}=S_{\triangle OBC}$,求点$M$的坐标.
(1) 求反比例函数与一次函数的解析式;
(2) 若点$M$在$x$轴上$,S_{\triangle OAM}=S_{\triangle OBC}$,求点$M$的坐标.
答案：23.(1)
∵点A(-1,6)在反比例函数的图象上，
∴k₂ = -1×6 = -6.
∴反比例函数的解析式为$y = -\frac{6}{x}.$将$B(a,2 - \frac{12}{a})$代入$y = -\frac{6}{x},$得$2 - \frac{12}{a} = -\frac{6}{a}，$解得a = 3.经检验，a = 3是原分式方程的解.
∴B(3,-2).将A(-1,6),B(3,-2)代入y = k₁x + b，得$\begin{cases}-k₁ + b = 6 \\ 3k₁ + b = -2\end{cases}，$解得$\begin{cases}k₁ = -2 \\ b = 4\end{cases}.$
∴一次函数的解析式为y = -2x + 4 (2)对于y = -2x + 4,令y = 0,则x = 2,
∴C(2,0).
∴OC = 2.
∴$S_△BOC = \frac{1}{2}OC·$|y_B|$ = \frac{1}{2}×2×2 = 2.$设点M的坐标为(m,0).
∵A(-1,6),S_△OAM = S_△OBC,
∴$\frac{1}{2}×6×$|m| = 2,解得$m = ±\frac{2}{3}.$
∴点M的坐标为$(\frac{2}{3},0)$或$(-\frac{2}{3},0)$
∵点A(-1,6)在反比例函数的图象上，
∴k₂ = -1×6 = -6.
∴反比例函数的解析式为$y = -\frac{6}{x}.$将$B(a,2 - \frac{12}{a})$代入$y = -\frac{6}{x},$得$2 - \frac{12}{a} = -\frac{6}{a}，$解得a = 3.经检验，a = 3是原分式方程的解.
∴B(3,-2).将A(-1,6),B(3,-2)代入y = k₁x + b，得$\begin{cases}-k₁ + b = 6 \\ 3k₁ + b = -2\end{cases}，$解得$\begin{cases}k₁ = -2 \\ b = 4\end{cases}.$
∴一次函数的解析式为y = -2x + 4 (2)对于y = -2x + 4,令y = 0,则x = 2,
∴C(2,0).
∴OC = 2.
∴$S_△BOC = \frac{1}{2}OC·$|y_B|$ = \frac{1}{2}×2×2 = 2.$设点M的坐标为(m,0).
∵A(-1,6),S_△OAM = S_△OBC,
∴$\frac{1}{2}×6×$|m| = 2,解得$m = ±\frac{2}{3}.$
∴点M的坐标为$(\frac{2}{3},0)$或$(-\frac{2}{3},0)$